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saveliy_v [14]
3 years ago
8

A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos

ition of fluorine
Chemistry
1 answer:
Volgvan3 years ago
4 0

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

percent composition element A=\frac{total mass of element A}{mass of compound} *100

In this case, the percent composition of fluorine is:

percent composition of fluorine=\frac{39.6 g}{60.3 g} *100

percent composition of fluorine= 65.67%

<u><em>The percent composition of fluorine is 65.67%</em></u>

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3 years ago
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<u>Answer:</u> The given chemical reaction can be classified as synthesis and exothermic.

<u>Explanation:</u>

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4 0
2 years ago
What mass of octane must be burned in order to liberate 5270 kj of heat? δhcomb = -5471 kj/mol?
katrin2010 [14]
As,
                           5471 kJ heat is given by  =  1 mole of Octane
Then,
                    5310 kJ heat will be given by  = X moles of Octane

Solving for X,
                                  X  =  (5310 kJ × 1 mol) ÷ 5471 kJ

                                  X  =  0.970 moles of Ocatne

So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
                                  Moles  =  mass / M.mass
Or,
                                  Mass  =  Moles × M.mass
Putting values,
                                  Mass  =  0.970 mol × 114.23 g/mol

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3 years ago
What is the predominant intermolecular force in CBr4
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Ilia_Sergeevich [38]

Explanation:

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moles × molar mass = mass

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3 0
2 years ago
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